An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #55 KJ# to # 42KJ# over #t in [0,4s]#. What is the average speed of the object?

Question

1240 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-30T09:33:18

The average speed is #=139.2ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=5kg#

The initial velocity is #=u_1ms^-1#

The final velocity is #=u_2 ms^-1#

The initial kinetic energy is #1/2m u_1^2=55000J#

The final kinetic energy is #1/2m u_2^2=42000J#

Therefore,

#u_1^2=2/5*55000=22000m^2s^-2#

and,

#u_2^2=2/5*42000=16800m^2s^-2#

The graph of #v^2=f(t)# is a straight line

The points are #(0,22000)# and #(4,16800)#

The equation of the line is

#v^2-22000=(16800-22000)/4t#

#v^2=-1300t+22000#

So,

#v=sqrt(-1300t+22000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4(sqrt(-1300t+22000))dt#

#4 barv=[((-1300t+22000)^(3/2)/(-3/2*1300))]_0^4#

#=((-1300*4+22000)^(3/2)/(-1950))-((-1300*0+22000)^(3/2)/(-1950))#

#=22000^(3/2)/1950-16800^(3/2)/1950#

#=556.7#

So,

#barv=556.7/4=139.2ms^-1#

The average speed is #=139.2ms^-1#